An adult weighing $600 \, N$ raises the centre of gravity of his body by $0.25 \, m$ while taking each step of $1 \, m$ length in jogging. If he jogs for $6 \, km$,calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $10 \, \%$ of energy intake in the form of food,calculate the energy equivalents of food that would be required to compensate for the energy utilised for jogging.

(N/A) Given: Weight of adult $W = mg = 600 \, N$,height raised per step $h = 0.25 \, m$,step length $l = 1 \, m$,total distance $d = 6 \, km = 6000 \, m$.
Total number of steps $n = \frac{d}{l} = \frac{6000 \, m}{1 \, m} = 6000$.
Energy utilised per step $E_{step} = mgh = 600 \, N \times 0.25 \, m = 150 \, J$.
Total energy utilised in jogging $E_{total} = n \times E_{step} = 6000 \times 150 \, J = 9 \times 10^5 \, J$.
Given that the body converts $10 \, \%$ of food energy intake into work,let $E_{food}$ be the total energy intake.
$0.10 \times E_{food} = 9 \times 10^5 \, J$.
$E_{food} = \frac{9 \times 10^5}{0.10} = 9 \times 10^6 \, J$.